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\begin{document}

\title{高等代数一}
\subtitle{11-矩阵的逆阵-伴随矩阵}
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年10月27日} }

\maketitle

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\begin{frame}{内容提要 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}
\item  矩阵的逆阵的定义
\item  伴随矩阵的定义
\item  用伴随矩阵求逆阵
\item  用初等变换方法求逆阵

\end{enumerate}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{11.1. 逆阵的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}

\item {\color{red}定义：设 $A$ 是 $n\times n$ 矩阵，若存在 $n\times n$ 矩阵 $B$, 使得 $AB=E_n$ 以及 $BA=E_n$ 都成立，则称矩阵 $B$ 是矩阵 $A$ 的逆阵，记为 $B=A^{-1}$. 这时称 $A$ 是可逆矩阵、非奇异矩阵、非退化矩阵等。}

\item 例子1：设 $A$ 是一个实数。则一阶矩阵 $A$ 可逆当且仅当 $A\neq 0$.

\item 例子2：设 {\footnotesize $A=\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}$}, 求逆阵 $A^{-1}$.

\item 解答：设逆阵为\, {\footnotesize $B=\begin{pmatrix}x&y \\ u&v \end{pmatrix}$}, 则有 $AB=E_2$ 以及 $BA=E_2$. 即 
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}
\begin{pmatrix}x&y \\ u&v \end{pmatrix}
=
\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix} \,\,\text{以及}\,\,
\begin{pmatrix}x&y \\ u&v \end{pmatrix}
\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}
=
\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

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\begin{itemize}

\item  我们先来看 $AB=E_2$. 按照矩阵乘法，可得
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix}x+2u&y+2v \\ 3x+4u&3y+4v \end{pmatrix}
=\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}.
\end{eqnarray*}
}
\item 因为矩阵相等是指分量对应相等，所以有
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x+2u &=& 1 \\
y+2v &=& 0 \\
3x+4u &=& 0 \\
3y+4v &=& 1\\
\end{array}\right.
\Rightarrow
\left\{\begin{array}{rcl}
x &=& -2 \\
y &=& 1 \\
u &=& 3/2 \\
v &=& -1/2\\
\end{array}\right.
\Rightarrow
B=\begin{pmatrix}-2&1 \\ 3/2&-1/2 \end{pmatrix}. 
\end{eqnarray*}
}
\item 最后验证 $BA=E_2$. 计算可得 {\footnotesize $BA=
\begin{pmatrix}-2&1 \\ 3/2&-1/2 \end{pmatrix}
\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}
=\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}=E_2
$. 
}

\end{itemize}

\end{frame}

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\begin{frame}{11.3. 伴随矩阵的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}
\item  {\color{red}定义：设 $A=(a_{ij})$ 是一个 $n\times n$ 矩阵。记 $A_{ij}$ 的 是元素 $a_{ij}$ 对应的代数余子式。则称矩阵 $A^*=(A_{ji})$ 为矩阵 $A$ 的伴随矩阵。}

\item  例子3：设 {\footnotesize $A=\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}$}, 求伴随矩阵 $A^*$.

\vspace{-0.2cm}

\item  解答：按伴随矩阵的定义，有
{\footnotesize $ A^* = \begin{pmatrix}A_{11}&A_{21} \\ A_{12}&A_{22} \end{pmatrix}
=\begin{pmatrix}4&-2 \\ -3&1 \end{pmatrix}$. }

\item  例子4：设 {\footnotesize $A=\begin{pmatrix}1&2&3 \\ 3&4&5 \\ 5&6&7 \end{pmatrix}$}, 求伴随矩阵 $A^*$.

\vspace{-0.2cm}

\item  解答：计算所有代数余子式可得
{\footnotesize $ A^* = \begin{pmatrix}A_{11}&A_{21}&A_{31} \\ A_{12}&A_{22}&A_{32} \\ A_{13}&A_{23}&A_{33} \end{pmatrix}
=\begin{pmatrix}-2&4&-2 \\ 4&-8&4 \\ -2&4&-2 \end{pmatrix}$. }

\end{itemize}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{11.4. 伴随矩阵与逆阵}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}

\item  {\color{red}定理：若矩阵 $A$ 的行列式的值不等于零，即 $\det(A)\neq 0$, 则 $A$ 的逆阵为
\begin{eqnarray*}
A^{-1} = \frac{1}{\det(A)}A^*.
\end{eqnarray*}
}
\item 证明来自另一个可能更常用的等式，直接计算验证下述等式，%记 $d=\det(A)$, 
{\footnotesize 
\begin{eqnarray*}
AA^* = \det(A)E_n = 
\begin{pmatrix}
\det(A)&0&\cdots &0 \\ 
0&\det(A)&\cdots &0 \\ 
\vdots & \vdots & \ddots &\vdots &\\ 
0&0&\cdots&\det(A) 
\end{pmatrix}.
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}
\item  例子5：设 {\footnotesize $A=\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}$}, 验证 $AA^* = \det(A)E_n$, 并由此写出 $A^{-1}$. 

\item  解答：
{\footnotesize 
\begin{eqnarray*}
AA^* &=& 
\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}
\begin{pmatrix}4&-2 \\ -3&1 \end{pmatrix}
=\begin{pmatrix}-2&0 \\ 0&-2 \end{pmatrix}=-2E_2, \\ 
\det{A} &=& \begin{vmatrix}1&2 \\ 3&4 \end{vmatrix} = -2, \\ 
A^{-1} &=& \frac{1}{\det(A)}A^* = \frac{1}{-2}\begin{pmatrix}4&-2 \\ -3&1 \end{pmatrix}
 = \begin{pmatrix}-2&1 \\ 3/2&-1/2 \end{pmatrix}. 
\end{eqnarray*}. 
}

\end{itemize}

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\begin{itemize}
\item  例子6：设 $A=(a_{ij})$ 是三阶矩阵，证明 $AA^* = \det(A)E_3$. 

\item  解答：
\begin{enumerate}
\item  记 $d=\det(A)$. 要证明的等式为
{\footnotesize 
\begin{eqnarray*}
AA^* =
\begin{pmatrix}
{\color{blue}a_{11}}&{\color{blue}a_{12}}&{\color{blue}a_{13}} \\ 
{\color{red}a_{21}}&{\color{red}a_{22}}&{\color{red}a_{23}} \\ 
a_{31}&a_{32}&a_{33} \\  
\end{pmatrix}
\begin{pmatrix}
{\color{blue}A_{11}}&A_{21}&{\color{red}A_{31}} \\ 
{\color{blue}A_{12}}&A_{22}&{\color{red}A_{32}} \\ 
{\color{blue}A_{13}}&A_{23}&{\color{red}A_{33}} 
\end{pmatrix}
=
\begin{pmatrix}{\color{blue}d}&0&0 \\ 0&d&{\color{red}0} \\ 0&0&d \end{pmatrix}. 
\end{eqnarray*}. 
}
\item  对角线上的三个 $d$ 正好是行列式 $|A|$ 按行展开。

\item  对角线之外的六个零正好是行列式 $|A|$ 中的某两行改成了相等。

\end{enumerate}


\end{itemize}

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\begin{frame}{11.7. 用初等变换方法求逆阵}

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\begin{itemize}
\item  例子7：设矩阵 {\footnotesize $A=\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}$}, 使用初等变换的方法，求 $A$ 的逆阵。

\item  解答：我们要找矩阵 $X$ 使得 $AX=E$. 我们的方法是对 $2\times 4$ 矩阵 $(A,E)$ 做{\color{red}行初等变换}，将 $A$ 所在的位置变成单位矩阵，这时原来 $E$ 所在的位置就自动成为了矩阵 $A$ 的逆阵。
{\footnotesize 
\begin{eqnarray*}
(A,E) = \begin{pmatrix}1&2 &1&0 \\ 3&4&0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第1行乘以$(-3)$加到第2行 }}& 
\begin{pmatrix}1&2&1&0 \\ 0&-2&-3&1 \end{pmatrix} \\
&\xrightarrow[\text{ }]{\text{第2行乘以$(-1/2)$ }}&
\begin{pmatrix}1&2&1&0 \\ 0&1&3/2&-1/2 \end{pmatrix} \\
&\xrightarrow[\text{ }]{\text{第2行乘以$(-2)$加到第1行 }}&
\begin{pmatrix}1&0&-2&1 \\ 0&1&3/2&-1/2 \end{pmatrix}
=(E,A^{-1}). 
\end{eqnarray*}
}

%所以矩阵 $A$ 的逆阵为 $A^{-1}=\begin{pmatrix}-2&1 \\ 3/2&-1/2 \end{pmatrix}$.

\end{itemize}

\end{frame}

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\begin{frame}{11.8. 用初等变换方法解矩阵方程}

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\begin{itemize}

\item  例子8：求解矩阵方程\, 
{\footnotesize $\begin{pmatrix}1&0 \\ 3&1 \end{pmatrix} X
=\begin{pmatrix}5&6\\ 7&8 \end{pmatrix}
$. } 这个方程记为 $AX=B$. 

\item  解答：对``增广矩阵'' $(A,B)$ 做{\color{red}行初等变换}，化为 $(E,A^{-1}B)$. 

{\footnotesize 
\begin{eqnarray*}
(A,B) = \begin{pmatrix}1&0 &5&6 \\ 3&1&7&8 \end{pmatrix}
\xrightarrow[\text{ }]{\text{第1行乘以$(-3)$加到第2行 }}
\begin{pmatrix}1&0&5&6 \\ 0&1&-8&-10 \end{pmatrix} = (E,A^{-1}B). 
\end{eqnarray*}
}

所以\,\, {\footnotesize $X=A^{-1}B = \begin{pmatrix}5&6 \\ -8&-10 \end{pmatrix}$}.

\end{itemize}

\end{frame}

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\begin{itemize}

\item  例子9：求解矩阵方程\, 
{\footnotesize $
X\begin{pmatrix}1&0 \\ 3&1 \end{pmatrix}
=\begin{pmatrix}5&6\\ 7&8 \end{pmatrix}
$}. 这个方程记为 $XA=B$. 

\item  解答：对``增广矩阵'' {\footnotesize $\begin{pmatrix}A \\ B \end{pmatrix}$}
 做{\color{red}列初等变换}，化为 {\footnotesize $\begin{pmatrix}E \\ BA^{-1} \end{pmatrix}$}. 
 
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix}A \\ B \end{pmatrix} 
= 
\begin{pmatrix}1&0 \\ 3&1\\ 5&6 \\ 7&8 \end{pmatrix}
\xrightarrow[\text{ }]{\text{第2列乘以$(-3)$加到第1列 }}
\begin{pmatrix}1&0 \\ 0&1 \\ -13&6 \\ -17&8 \end{pmatrix} 
= 
\begin{pmatrix}E \\ BA^{-1} \end{pmatrix}. 
\end{eqnarray*}
}

所以\,\, {\footnotesize $X=BA^{-1} = \begin{pmatrix} -13&6 \\ -17&8 \end{pmatrix}$}.



\end{itemize}

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\begin{frame}{11.10. 用初等变换方法求解矩阵方程的原理}

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\begin{itemize}

\item  用行初等变换实现 $AX=B \Rightarrow X=A^{-1}B$ 的原理：
\begin{enumerate}
\item  对矩阵 $A$ 做{\color{red}行初等变换}等价于在矩阵 $A$ 的{\color{red}左边乘以相应的初等矩阵}。
\item  由 $AX=B$ 可得 $P(AX)=PB$. 
\item  根据矩阵乘法的结合律，可得 $(PA)X=PB$. 
\item  选取一些初等矩阵 $P_1,P_2,\cdots, P_s$ 使得 $P_s\cdots P_2P_1A=E$. 
\item  可得 $X=P_s\cdots P_2P_1B$.  
\end{enumerate}

\item  用列初等变换实现 $XA=B \Rightarrow X=BA^{-1}$ 的原理：
\begin{enumerate}
\item  对矩阵 $A$ 做{\color{red}列初等变换}等价于在矩阵 $A$ 的{\color{red}右边乘以相应的初等矩阵}。
\item  由 $XA=B$ 可得 $(XA)P=BP$. 
\item  根据矩阵乘法的结合律，可得 $X(AP)=BP$. 
\item  选取一些初等矩阵 $P_1,P_2,\cdots, P_s$ 使得 $AP_1P_2\cdots P_s=E$. 
\item  可得 $X=BP_1P_2\cdots P_s$.  
\end{enumerate}


\end{itemize}

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\begin{enumerate}

\item  考虑矩阵\,  {\footnotesize $A=\begin{pmatrix}1&2&-1 \\ 3&1&0 \\ -1&0&-2 \end{pmatrix}$}. 
\begin{enumerate}
\item  计算伴随矩阵 $A^*$, 并验证 $AA^*=\det(A)E$. 
\item  使用初等变换的方法，求 $A$ 的逆阵。
\end{enumerate}

\item  求解矩阵方程\, {\footnotesize $ \begin{pmatrix}2&0 \\ 3&1 \end{pmatrix} X =\begin{pmatrix}4&6\\ 5&7 \end{pmatrix} $. } 

\end{enumerate}

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\begin{enumerate}

\item  
\begin{enumerate}

\item  伴随矩阵为\, {\footnotesize $A^*=\begin{pmatrix} -2&4&1 \\  6&-3&-3 \\  1&-2&-5 \end{pmatrix}$}, 
计算可得\, {\footnotesize $AA^*=\begin{pmatrix} 9&0&0 \\  0&9&0 \\  0&0&9 \end{pmatrix}$}. 

\item 逆阵为\, {\footnotesize $A^{-1}=\begin{pmatrix} -2/9&4/9&1/9 \\  6/9&-3/9&-3/9 \\  1/9&-2/9&-5/9 \end{pmatrix}$}. 

\end{enumerate}

\item  所求矩阵为\, {\footnotesize $X =\begin{pmatrix}2&3\\ -1&-2 \end{pmatrix} $. } 

\end{enumerate}

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